What does it mean to integrate a Brownian motion with respect to time? I am reading about stochastic process, but could not make sense if one equation I encountered. Can anyone help me understand it?
The equation states that suppose R(s) is an interest rate process, then the discount process is $D(t)=e^{- \int_0^t R(s)ds} $.
Suppose R(t)=W(t) is a simple Brownian motion, what does $\int_0^t R(s)ds$ mean?
Is it a Lebesgue integral? Or is it an Ito's integral? How to interpret it intuitively?
This is from chapter 5 of Shreve's stochastic calculus for finance, equation 5.2.17 on page 215.
 A: It might be easier to think about if we abstract a little.  Brownian motion is just a continuous, time dependent random variable determined by some probability space $(X,\mathcal F, P)$.  Let us view it as a function of two variables, $f(x,t)$.  If $x$ is fixed, we have a continuous function, $f_x(t)=f(x,t)$, and we can compute it's integral $F(x,t)=F_x(t)=\int_0^t f_x(s) ds$, where we have just taken the usual integral of a continuous function (Riemann, Lebesgue, whatever, all the definitions agree for nice continuous functions).  Thus, we have a transformation of the space of time dependent continuous random variables on $X$ (i.e., functions on $X\times \mathbb R$).  The fact that we are working in particular with brownian motion doesn't enter into things.
A more complicated question is what it means to integrate a function or a random variable with respect to Brownian motion.  This gets you to the Ito integral (and other similar variants) which are more subtle.
A: Since Brownian Motion has continuous paths with probability 1 thus$\int_{0}^{t} W_s ds$ is well define. Now set $f(x)=x^3$. By application of Ito's lemma, we have
$$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds$$
$$W^3(t)=3\int_{0}^{t}W^2_s\,dW_s+3\int_{0}^{t}W_sds$$
therefore
$$\int_{0}^{t}W_sds=\frac{1}{3}W^3(t)-\int_{0}^{t}W^2_s\,dW_s$$
Note 
The Itô integral can be defined in a manner similar to the Riemann–Stieltjes integral, that is as a limit in probability of Riemann sums; such a limit does not necessarily exist pathwise. Suppose that $W_t$ is a Wiener process and that $X_t$ is a right-continuous (cadlag), adapted and locally bounded process if $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$  with mesh going to zero, then the Itô integral of $X_t$ with respect to $W_t$ up to time t is a random variable
$$\int_{0}^{t}X_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{X({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$
Set $X_s=W^2_s$, we  have
$$\int_{0}^{t}W^2_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{W^2({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$
Edit
\begin{eqnarray}
\int_{0}^t W_s \,ds & = & \int_0^t \int_0^t \mathbf{1}_{[0,s]} (u) \,dW_u \,ds \\
& = & \int_0^t \int_0^t \mathbf{1}_{[0,s]} (u) \,ds  \,dW_u\\
& = &  \int_0^t (t-u) \,dW_u\\
\end{eqnarray}
Indeed
$$\int_{0}^{t}{{{W}_{s}}}ds\sim N\left( 0\,,\,\frac{1}{3}{{t}^{3}} \right)$$
A: There are two types of integrals involving Brownian motion, time integral and itos integral. Here the integral is a time integral, which is just an ordinary Lebesgue integral. The integral itself is also a random variable as it depends on the path of brownian motion. 
The interpretation and properties of a time integral should follow those of Lebesgue integrals. 
